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Suppose $f(x)$ is continuous on $[0,\,1]$, differentiable in $(0,\,1)$, and $$ f(0)=f(1)=0,\qquad f(\frac{1}{2})=1. $$ Show that $\exists\;\xi\in(0,\,1)$, such that $$ f'(\xi)-3\big(f(\xi)-\xi\big)=1. $$

My idea is to get a primitive of $f'(t)-3\big(f(t)-t\big)=1$ and use Rolle theorem. But I don't know how to use $f(\frac{1}{2})=1$.

Knt
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    Look at the function $g(x)=(f(x)-x)e^{-3x}$. It takes the values $g(0)=0$, $g(1/2)=e^{-3/2}/2>0$ and $g(1)=-e^{3}<0$. Therefore, there are two different points in $[0,1]$ where it takes the same value. By Rolle's theorem, in between its derivative vanishes somewhere $c$. Then $0=g'(c)=e^{-3c}[(f'(c)-1)-3(f(c)-c)]$. Since $e^{-3c}\neq0$, it follows that $(f'(c)-1)-3(f(c)-c)=0$. – conditionalMethod Nov 09 '19 at 10:52

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